complex numbrs (1 Viewer)

[sssona09]

show that z^5-1/ z-1 = z^4 +z^3 +z^2 +z+1

also show that z^5-1/z-1= (z^2-2zcos2pi/5+1)(z^2-2zcos4pi/5+1)

 

[fluffchuck]

Show that (z^5 - 1)/(z - 1) = z^4 + z^3 + z^2 + z + 1
Start off with your right hand side and consider the sum of a GP.

Also show that (z^5 - 1)/(z - 1) = (z^2 - 2zcos(2pi/5) + 1)(z^2 - 2zcos(4pi/5) + 1)
On the left hand side, we have (z^5 - 1)/(z-1). Try to factorise (z^5 - 1) into the form (z - a)(z - b)(z - c)(z - d)(z - e) where a, b, c, d and e are the zeros of (z^5 - 1) (note: to obtain the zeros of (z^5 - 1), we just set (z^5 - 1) to 0, in other words, z^5 - 1 = 0, which is z^5 = 1, which is root of unity!). From here, you should be able to apply the identity , and hence, obtain your right hand side. I hope this helps